Multiply both sides of the second equation by 100 to get rid of the decimals:
0.05<em>n</em> + 0.10<em>d</em> = 1.50
==> 5<em>n</em> + 10<em>d</em> = 150
Multiply both sides of the first equation by -5:
<em>n</em> + <em>d</em> = 21
==> -5<em>n</em> - 5<em>d</em> = -105
Add the two equations together:
(5<em>n</em> + 10<em>d</em>) + (-5<em>n</em> - 5<em>d</em>) = 150 + (-105)
Notice that the terms containing <em>n</em> get eliminated and we can solve for <em>d</em> :
(5<em>n</em> - 5<em>n</em>) + (10<em>d</em> - 5<em>d</em>) = 150 - 105
5<em>d</em> = 45
<em>d</em> = 45/5 = 9
Plug this into either original equation to solve for <em>n</em>. Doing this with the first equation is easiest:
<em>n</em> + 9 = 21
<em>n</em> = 21 - 9 = 12
So Donna used 12 nickels and 9 dimes.
Answer:
- x = 0 or 1
- x = ±i/4
- x = -5 (twice)
Step-by-step explanation:
Factoring is aided by having the equations in standard form. The first step in each case is to put the equations in that form. The zero product property tells you that a product is zero when a factor is zero. The solutions are the values of x that make the factors zero.
1. x^2 -x = 0
x(x -1) = 0 . . . . . x = 0 or 1
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2. 16x^2 +1 = 0
This is the "difference of squares" ...
(4x)^2 - (i)^2 = 0
(4x -i)(4x +i) = 0 . . . . . x = -i/4 or i/4 (zeros are complex)
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3. x^2 +10x +25 = 0
(x +5)(x +5) = 0 . . . . . x = -5 with multiplicity 2
Answer:
b is answer?
Step-by-step explanation:
feels right
.
I converted it into decimal form with a calculator and got approximately 48.9898
And if you round that, you will get the 49, which fits the inequality of
45<
< 50 since it becomes 45 < 49 < 50 which is true